Question: Define $f(x, y, z) = x^2 - 6y + 2z$. Let $\vec{a} = (1, -1, 1)$ and $\vec{v} = \left( -1, 2, 2 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Answer: Our ultimate goal is to substitute $h = 0$ into the limit and get the answer. To do this, we first need to cancel out the $h$ in the denominator. Otherwise we're dividing by zero. How can we simplify the limit so that we get an $h$ on the top and an $h$ on the bottom, which we can then cancel? Let's plug in $\vec{a}$ and $\vec{v}$ to the limit. $ \lim_{h \to 0} \dfrac{f \left( (1, -1, 1) + h \left( -1, 2, 2 \right) \right) - f(1, -1, 1)}{h}$ Now we can add together the vectors. $ \lim_{h \to 0} \dfrac{f \left( 1 - h, -1 + 2h, 1 + 2h \right) - f(1, -1, 1)}{h}$ Let's evaluate $f$. $ \lim_{h \to 0} \dfrac{(1 - h)^2 - 6(-1 + 2h) + 2(1 + 2h) - (1^2 -6(-1) + 2)}{h}$ We can cancel out all the terms without an $h$. $ \lim_{h \to 0} \dfrac{1 - 2h + h^2 + 6 - 12h + 2 + 4h - 1 - 6 - 2}{h}$ becomes $ \lim_{h \to 0} \dfrac{h^2 - 10h}{h}$ Now we can cancel $h$ and calculate the limit. $\begin{aligned} \lim_{h \to 0} \dfrac{h^2 - 10h}{h} &= \lim_{h \to 0} h - 10 \\ \\ &= -10 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = -10$.